Calculate the concentration of NH4+ from ammonium chloride required to prevent the precipitation of Ca(OH)2 in a liter of solution that contains .10 mol of ammonia ans .10 mol pf calcium ion.
I got the two equations but I do not know where to begin.
The question asks for (NH4^+) yet you give an answer in Ca ion which doesn't make sense. Could you have meant that the solution contains 0.1 mol Ca ion?
yes .1 M Ca ion and .1 M ammonia
Ksp Ca(OH)2 = (Ca^+2)(OH^-)^2 =??
Kb = (NH4^+)(OH^-)/(NH3) =??
From equation 1, calculate (OH^-) needed for pptn with 0.1 M Ca ion.
From equation 2, using 0.1 M for (NH3), and the (OH^-) from equation 1, calculate (NH4^+).
Post your work if you get stuck.
N9ooooooo
To calculate the concentration of NH4+ from ammonium chloride required to prevent the precipitation of Ca(OH)2, we can follow these steps:
Step 1: Write down the relevant equations:
The solubility product constant, Ksp, for Ca(OH)2 is given by:
Ksp = [Ca^+2][OH^-]^2
The base dissociation constant, Kb, for the ammonium ion, NH4+, is given by:
Kb = ([NH4^+][OH^-])/[NH3]
Step 2: Calculate (OH^-) needed for precipitation with 0.1 M Ca ion:
Since the concentration of Ca ion is 0.1 M and the solubility product constant is known, we can rearrange the equation to solve for (OH^-):
Ksp = [Ca^+2][OH^-]^2
0.1 = (OH^-)^2
(OH^-) = sqrt(0.1)
Step 3: Calculate (NH4^+):
Substituting the calculated value of (OH^-) into the base dissociation constant equation, we have:
Kb = ([NH4^+][OH^-])/[NH3]
Using the given values of Kb, [NH3], and (OH^-), we can solve for [NH4^+].
Make sure to fill in the actual values and perform the calculations. If you encounter any specific issues during the calculation or have any additional questions, feel free to ask for further assistance.